Riemann Zeta function is Big Oh of log(|t|)

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I'm stuck proving the following:



for all $t$ such that $|t|$ sufficiently large and if $sigma > 1 - frac100log $ $zeta(sigma + it) = O(log|t|)$



Questions:



  • Do we only consider the real part and if so why?

  • In what we should obtain (below) how do we obtain the |t| in the denominator of the first error term?

  • In what we should obtain (below) how do we get the $|sigma + it|$ in the error term, the other part clearly follows by computing the integral

  • How are these error terms O(1)?

Attempt:



Letting $x=|t|$ then by definition of $zeta(s) = sum_n leq frac1n^s + fracts-1 + frac^s - s int_^infty w fracdww^s+1$



After substituting in for $sigma$ (rewriting $s = sigma + it$) I obtain



$$ sum_n leq frac1n^sigma + it + frac-frac100-it + frac^sigma + it - s int_^infty w fracdww^s+1 $$



I should somehow obtain



$$zeta(sigma + it) = sum_n leq frac1n^sigma + it + O(fract) + O(|sigma + it||t|^- sigma)$$
$$ = sum_n leq frac1n^sigma + it + O(1)$$



Thanks.




proof-explanation analytic-number-theory riemann-zeta




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    up vote
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    down vote

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    I'm stuck proving the following:



    for all $t$ such that $|t|$ sufficiently large and if $sigma > 1 - frac100log $ $zeta(sigma + it) = O(log|t|)$



    Questions:



    • Do we only consider the real part and if so why?

    • In what we should obtain (below) how do we obtain the |t| in the denominator of the first error term?

    • In what we should obtain (below) how do we get the $|sigma + it|$ in the error term, the other part clearly follows by computing the integral

    • How are these error terms O(1)?

    Attempt:



    Letting $x=|t|$ then by definition of $zeta(s) = sum_n leq frac1n^s + fracts-1 + frac^s - s int_^infty w fracdww^s+1$



    After substituting in for $sigma$ (rewriting $s = sigma + it$) I obtain



    $$ sum_n leq frac1n^sigma + it + frac-frac100-it + frac^sigma + it - s int_^infty w fracdww^s+1 $$



    I should somehow obtain



    $$zeta(sigma + it) = sum_n leq frac1n^sigma + it + O(fract) + O(|sigma + it||t|^- sigma)$$
    $$ = sum_n leq frac1n^sigma + it + O(1)$$



    Thanks.




    proof-explanation analytic-number-theory riemann-zeta




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      up vote
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      down vote

      favorite









      up vote
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      down vote

      favorite











      I'm stuck proving the following:



      for all $t$ such that $|t|$ sufficiently large and if $sigma > 1 - frac100log $ $zeta(sigma + it) = O(log|t|)$



      Questions:



      • Do we only consider the real part and if so why?

      • In what we should obtain (below) how do we obtain the |t| in the denominator of the first error term?

      • In what we should obtain (below) how do we get the $|sigma + it|$ in the error term, the other part clearly follows by computing the integral

      • How are these error terms O(1)?

      Attempt:



      Letting $x=|t|$ then by definition of $zeta(s) = sum_n leq frac1n^s + fracts-1 + frac^s - s int_^infty w fracdww^s+1$



      After substituting in for $sigma$ (rewriting $s = sigma + it$) I obtain



      $$ sum_n leq frac1n^sigma + it + frac-frac100-it + frac^sigma + it - s int_^infty w fracdww^s+1 $$



      I should somehow obtain



      $$zeta(sigma + it) = sum_n leq frac1n^sigma + it + O(fract) + O(|sigma + it||t|^- sigma)$$
      $$ = sum_n leq frac1n^sigma + it + O(1)$$



      Thanks.




      proof-explanation analytic-number-theory riemann-zeta




      share|cite|improve this question











      I'm stuck proving the following:



      for all $t$ such that $|t|$ sufficiently large and if $sigma > 1 - frac100log $ $zeta(sigma + it) = O(log|t|)$



      Questions:



      • Do we only consider the real part and if so why?

      • In what we should obtain (below) how do we obtain the |t| in the denominator of the first error term?

      • In what we should obtain (below) how do we get the $|sigma + it|$ in the error term, the other part clearly follows by computing the integral

      • How are these error terms O(1)?

      Attempt:



      Letting $x=|t|$ then by definition of $zeta(s) = sum_n leq frac1n^s + fracts-1 + frac^s - s int_^infty w fracdww^s+1$



      After substituting in for $sigma$ (rewriting $s = sigma + it$) I obtain



      $$ sum_n leq frac1n^sigma + it + frac-frac100-it + frac^sigma + it - s int_^infty w fracdww^s+1 $$



      I should somehow obtain



      $$zeta(sigma + it) = sum_n leq frac1n^sigma + it + O(fract) + O(|sigma + it||t|^- sigma)$$
      $$ = sum_n leq frac1n^sigma + it + O(1)$$



      Thanks.





      proof-explanation analytic-number-theory riemann-zeta





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      asked Jul 16 at 19:24









      VBACODER

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          Do we only consider the real part and if so why?




          If you are referring to the exponents, then yes, we consider only the real part. The reason is that we are using the real-valued logarithm of $a > 0$ to define $a^s = exp (scdot log a)$, and $lvert exp wrvert = exp (operatornameRe w)$ (since $lvert exp wrvert^2 = (exp w)(overlineexp w) = (exp w)(exp overlinew) = exp (w + overlinew) = exp(2operatornameRe w)$). So $lvert a^s rvert = a^sigma$ for $a > 0$.




          In what we should obtain (below) how do we obtain the $lvert trvert$ in the denominator of the first error term?




          The denominator is $s-1$, and we have $lvert s-1rvert geqslant lvert operatornameIm (s-1)rvert = lvert trvert$. This yields the estimate
          $$biggllvert fraclvert trvert^1-ss-1biggrrvert = fraclvert trvert^1-sigmalvert s-1rvert leqslant fraclvert trvert^1-sigmalvert trvert = lvert trvert^-sigma < lvert trvert^frac100log lvert trvert - 1 = frace^100lvert trvert$$
          for $lvert trvert > 1$, since then $lvert trvert^-sigma$ is decreasing in $sigma$ and $lvert trvert^b/log lvert trvert = exp bigl(fracblog lvert trvertcdot log lvert trvertbigr) = e^b$.



          The next term is fairly direct,
          $$biggllvert fraclbrace lvert trvertrbracelvert trvert^sbiggrrvert = fraclbracelvert trvertrbracelvert trvert^sigma < frac1lvert trvert^sigma < frace^100lvert trvert$$
          for $lvert trvert geqslant 1$ since $0 leqslant lbrace lvert trvertrbrace < 1$.



          For the integral term, we cannot go quite so far. The integral only makes sense for $operatornameRe s > 0$ (perhaps also for $operatornameRe s = 0$ and $sneq 0$, but not beyond that), so we must either restrict to $lvert trvert geqslant e^100$ or modify the lower bound for $sigma$ to
          $$sigma > max ; Biggl 0, 1 - frac100log lvert trvertBiggr,.$$
          Well, since the straightforward estimate of the integral,
          $$Biggllvert int_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant int_lvert trvert^infty fraclbrace wrbracew^1+sigma,dw leqslant int_lvert trvert^infty fracdww^1+sigma = frac1sigmacdot lvert trvert^sigma,,$$
          produces a $sigma$ in the denominator, we must keep $sigma$ away from zero, so we actually need a stronger restriction,
          $$sigma geqslant max; Biggl c, 1 - frac100log lvert trvertBiggr$$
          for a fixed $c in (0,1)$, or $lvert trvert geqslant t_0$ for a fixed $t_0 > e^100$.



          With that, we can estimate the last term by
          $$Biggllvert sint_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant fraclvert sigma + itrvertsigmalvert trvert^sigma leqslant fracsigma + lvert trvertsigmalvert trvert^sigma = lvert trvert^-sigma + fraclvert trvert^1-sigmasigma,.$$
          We had estimated $lvert trvert^-sigma$ above, and with that we obtain
          $$Biggllvert sint_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant frace^100lvert trvert + frace^100sigma leqslant e^100biggl(frac1lvert trvert + frac1cbiggr),.$$
          Thus each of the terms except the partial sum is bounded and we have
          $$zeta(sigma + it) = sum_n leqslant lvert trvert frac1n^sigma +it + O(1)$$
          in the indicated region. The remaining sum is estimated using
          $$lvert n^-sigma - itrvert = n^-sigma = fracn^1-sigman leqslant fraclvert trvert^1-sigman leqslant frace^100n$$
          for $sigma < 1$, and
          $$lvert n^-sigma-itrvert = n^-sigma leqslant frac1n$$
          for $sigma geqslant 1$.






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          • Thank you for such a detailed and clear response!
            – VBACODER
            Jul 26 at 19:32










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          1 Answer
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          active

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          up vote
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          down vote



          accepted











          Do we only consider the real part and if so why?




          If you are referring to the exponents, then yes, we consider only the real part. The reason is that we are using the real-valued logarithm of $a > 0$ to define $a^s = exp (scdot log a)$, and $lvert exp wrvert = exp (operatornameRe w)$ (since $lvert exp wrvert^2 = (exp w)(overlineexp w) = (exp w)(exp overlinew) = exp (w + overlinew) = exp(2operatornameRe w)$). So $lvert a^s rvert = a^sigma$ for $a > 0$.




          In what we should obtain (below) how do we obtain the $lvert trvert$ in the denominator of the first error term?




          The denominator is $s-1$, and we have $lvert s-1rvert geqslant lvert operatornameIm (s-1)rvert = lvert trvert$. This yields the estimate
          $$biggllvert fraclvert trvert^1-ss-1biggrrvert = fraclvert trvert^1-sigmalvert s-1rvert leqslant fraclvert trvert^1-sigmalvert trvert = lvert trvert^-sigma < lvert trvert^frac100log lvert trvert - 1 = frace^100lvert trvert$$
          for $lvert trvert > 1$, since then $lvert trvert^-sigma$ is decreasing in $sigma$ and $lvert trvert^b/log lvert trvert = exp bigl(fracblog lvert trvertcdot log lvert trvertbigr) = e^b$.



          The next term is fairly direct,
          $$biggllvert fraclbrace lvert trvertrbracelvert trvert^sbiggrrvert = fraclbracelvert trvertrbracelvert trvert^sigma < frac1lvert trvert^sigma < frace^100lvert trvert$$
          for $lvert trvert geqslant 1$ since $0 leqslant lbrace lvert trvertrbrace < 1$.



          For the integral term, we cannot go quite so far. The integral only makes sense for $operatornameRe s > 0$ (perhaps also for $operatornameRe s = 0$ and $sneq 0$, but not beyond that), so we must either restrict to $lvert trvert geqslant e^100$ or modify the lower bound for $sigma$ to
          $$sigma > max ; Biggl 0, 1 - frac100log lvert trvertBiggr,.$$
          Well, since the straightforward estimate of the integral,
          $$Biggllvert int_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant int_lvert trvert^infty fraclbrace wrbracew^1+sigma,dw leqslant int_lvert trvert^infty fracdww^1+sigma = frac1sigmacdot lvert trvert^sigma,,$$
          produces a $sigma$ in the denominator, we must keep $sigma$ away from zero, so we actually need a stronger restriction,
          $$sigma geqslant max; Biggl c, 1 - frac100log lvert trvertBiggr$$
          for a fixed $c in (0,1)$, or $lvert trvert geqslant t_0$ for a fixed $t_0 > e^100$.



          With that, we can estimate the last term by
          $$Biggllvert sint_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant fraclvert sigma + itrvertsigmalvert trvert^sigma leqslant fracsigma + lvert trvertsigmalvert trvert^sigma = lvert trvert^-sigma + fraclvert trvert^1-sigmasigma,.$$
          We had estimated $lvert trvert^-sigma$ above, and with that we obtain
          $$Biggllvert sint_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant frace^100lvert trvert + frace^100sigma leqslant e^100biggl(frac1lvert trvert + frac1cbiggr),.$$
          Thus each of the terms except the partial sum is bounded and we have
          $$zeta(sigma + it) = sum_n leqslant lvert trvert frac1n^sigma +it + O(1)$$
          in the indicated region. The remaining sum is estimated using
          $$lvert n^-sigma - itrvert = n^-sigma = fracn^1-sigman leqslant fraclvert trvert^1-sigman leqslant frace^100n$$
          for $sigma < 1$, and
          $$lvert n^-sigma-itrvert = n^-sigma leqslant frac1n$$
          for $sigma geqslant 1$.






          share|cite|improve this answer





















          • Thank you for such a detailed and clear response!
            – VBACODER
            Jul 26 at 19:32














          up vote
          1
          down vote



          accepted











          Do we only consider the real part and if so why?




          If you are referring to the exponents, then yes, we consider only the real part. The reason is that we are using the real-valued logarithm of $a > 0$ to define $a^s = exp (scdot log a)$, and $lvert exp wrvert = exp (operatornameRe w)$ (since $lvert exp wrvert^2 = (exp w)(overlineexp w) = (exp w)(exp overlinew) = exp (w + overlinew) = exp(2operatornameRe w)$). So $lvert a^s rvert = a^sigma$ for $a > 0$.




          In what we should obtain (below) how do we obtain the $lvert trvert$ in the denominator of the first error term?




          The denominator is $s-1$, and we have $lvert s-1rvert geqslant lvert operatornameIm (s-1)rvert = lvert trvert$. This yields the estimate
          $$biggllvert fraclvert trvert^1-ss-1biggrrvert = fraclvert trvert^1-sigmalvert s-1rvert leqslant fraclvert trvert^1-sigmalvert trvert = lvert trvert^-sigma < lvert trvert^frac100log lvert trvert - 1 = frace^100lvert trvert$$
          for $lvert trvert > 1$, since then $lvert trvert^-sigma$ is decreasing in $sigma$ and $lvert trvert^b/log lvert trvert = exp bigl(fracblog lvert trvertcdot log lvert trvertbigr) = e^b$.



          The next term is fairly direct,
          $$biggllvert fraclbrace lvert trvertrbracelvert trvert^sbiggrrvert = fraclbracelvert trvertrbracelvert trvert^sigma < frac1lvert trvert^sigma < frace^100lvert trvert$$
          for $lvert trvert geqslant 1$ since $0 leqslant lbrace lvert trvertrbrace < 1$.



          For the integral term, we cannot go quite so far. The integral only makes sense for $operatornameRe s > 0$ (perhaps also for $operatornameRe s = 0$ and $sneq 0$, but not beyond that), so we must either restrict to $lvert trvert geqslant e^100$ or modify the lower bound for $sigma$ to
          $$sigma > max ; Biggl 0, 1 - frac100log lvert trvertBiggr,.$$
          Well, since the straightforward estimate of the integral,
          $$Biggllvert int_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant int_lvert trvert^infty fraclbrace wrbracew^1+sigma,dw leqslant int_lvert trvert^infty fracdww^1+sigma = frac1sigmacdot lvert trvert^sigma,,$$
          produces a $sigma$ in the denominator, we must keep $sigma$ away from zero, so we actually need a stronger restriction,
          $$sigma geqslant max; Biggl c, 1 - frac100log lvert trvertBiggr$$
          for a fixed $c in (0,1)$, or $lvert trvert geqslant t_0$ for a fixed $t_0 > e^100$.



          With that, we can estimate the last term by
          $$Biggllvert sint_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant fraclvert sigma + itrvertsigmalvert trvert^sigma leqslant fracsigma + lvert trvertsigmalvert trvert^sigma = lvert trvert^-sigma + fraclvert trvert^1-sigmasigma,.$$
          We had estimated $lvert trvert^-sigma$ above, and with that we obtain
          $$Biggllvert sint_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant frace^100lvert trvert + frace^100sigma leqslant e^100biggl(frac1lvert trvert + frac1cbiggr),.$$
          Thus each of the terms except the partial sum is bounded and we have
          $$zeta(sigma + it) = sum_n leqslant lvert trvert frac1n^sigma +it + O(1)$$
          in the indicated region. The remaining sum is estimated using
          $$lvert n^-sigma - itrvert = n^-sigma = fracn^1-sigman leqslant fraclvert trvert^1-sigman leqslant frace^100n$$
          for $sigma < 1$, and
          $$lvert n^-sigma-itrvert = n^-sigma leqslant frac1n$$
          for $sigma geqslant 1$.






          share|cite|improve this answer





















          • Thank you for such a detailed and clear response!
            – VBACODER
            Jul 26 at 19:32












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted







          Do we only consider the real part and if so why?




          If you are referring to the exponents, then yes, we consider only the real part. The reason is that we are using the real-valued logarithm of $a > 0$ to define $a^s = exp (scdot log a)$, and $lvert exp wrvert = exp (operatornameRe w)$ (since $lvert exp wrvert^2 = (exp w)(overlineexp w) = (exp w)(exp overlinew) = exp (w + overlinew) = exp(2operatornameRe w)$). So $lvert a^s rvert = a^sigma$ for $a > 0$.




          In what we should obtain (below) how do we obtain the $lvert trvert$ in the denominator of the first error term?




          The denominator is $s-1$, and we have $lvert s-1rvert geqslant lvert operatornameIm (s-1)rvert = lvert trvert$. This yields the estimate
          $$biggllvert fraclvert trvert^1-ss-1biggrrvert = fraclvert trvert^1-sigmalvert s-1rvert leqslant fraclvert trvert^1-sigmalvert trvert = lvert trvert^-sigma < lvert trvert^frac100log lvert trvert - 1 = frace^100lvert trvert$$
          for $lvert trvert > 1$, since then $lvert trvert^-sigma$ is decreasing in $sigma$ and $lvert trvert^b/log lvert trvert = exp bigl(fracblog lvert trvertcdot log lvert trvertbigr) = e^b$.



          The next term is fairly direct,
          $$biggllvert fraclbrace lvert trvertrbracelvert trvert^sbiggrrvert = fraclbracelvert trvertrbracelvert trvert^sigma < frac1lvert trvert^sigma < frace^100lvert trvert$$
          for $lvert trvert geqslant 1$ since $0 leqslant lbrace lvert trvertrbrace < 1$.



          For the integral term, we cannot go quite so far. The integral only makes sense for $operatornameRe s > 0$ (perhaps also for $operatornameRe s = 0$ and $sneq 0$, but not beyond that), so we must either restrict to $lvert trvert geqslant e^100$ or modify the lower bound for $sigma$ to
          $$sigma > max ; Biggl 0, 1 - frac100log lvert trvertBiggr,.$$
          Well, since the straightforward estimate of the integral,
          $$Biggllvert int_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant int_lvert trvert^infty fraclbrace wrbracew^1+sigma,dw leqslant int_lvert trvert^infty fracdww^1+sigma = frac1sigmacdot lvert trvert^sigma,,$$
          produces a $sigma$ in the denominator, we must keep $sigma$ away from zero, so we actually need a stronger restriction,
          $$sigma geqslant max; Biggl c, 1 - frac100log lvert trvertBiggr$$
          for a fixed $c in (0,1)$, or $lvert trvert geqslant t_0$ for a fixed $t_0 > e^100$.



          With that, we can estimate the last term by
          $$Biggllvert sint_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant fraclvert sigma + itrvertsigmalvert trvert^sigma leqslant fracsigma + lvert trvertsigmalvert trvert^sigma = lvert trvert^-sigma + fraclvert trvert^1-sigmasigma,.$$
          We had estimated $lvert trvert^-sigma$ above, and with that we obtain
          $$Biggllvert sint_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant frace^100lvert trvert + frace^100sigma leqslant e^100biggl(frac1lvert trvert + frac1cbiggr),.$$
          Thus each of the terms except the partial sum is bounded and we have
          $$zeta(sigma + it) = sum_n leqslant lvert trvert frac1n^sigma +it + O(1)$$
          in the indicated region. The remaining sum is estimated using
          $$lvert n^-sigma - itrvert = n^-sigma = fracn^1-sigman leqslant fraclvert trvert^1-sigman leqslant frace^100n$$
          for $sigma < 1$, and
          $$lvert n^-sigma-itrvert = n^-sigma leqslant frac1n$$
          for $sigma geqslant 1$.






          share|cite|improve this answer














          Do we only consider the real part and if so why?




          If you are referring to the exponents, then yes, we consider only the real part. The reason is that we are using the real-valued logarithm of $a > 0$ to define $a^s = exp (scdot log a)$, and $lvert exp wrvert = exp (operatornameRe w)$ (since $lvert exp wrvert^2 = (exp w)(overlineexp w) = (exp w)(exp overlinew) = exp (w + overlinew) = exp(2operatornameRe w)$). So $lvert a^s rvert = a^sigma$ for $a > 0$.




          In what we should obtain (below) how do we obtain the $lvert trvert$ in the denominator of the first error term?




          The denominator is $s-1$, and we have $lvert s-1rvert geqslant lvert operatornameIm (s-1)rvert = lvert trvert$. This yields the estimate
          $$biggllvert fraclvert trvert^1-ss-1biggrrvert = fraclvert trvert^1-sigmalvert s-1rvert leqslant fraclvert trvert^1-sigmalvert trvert = lvert trvert^-sigma < lvert trvert^frac100log lvert trvert - 1 = frace^100lvert trvert$$
          for $lvert trvert > 1$, since then $lvert trvert^-sigma$ is decreasing in $sigma$ and $lvert trvert^b/log lvert trvert = exp bigl(fracblog lvert trvertcdot log lvert trvertbigr) = e^b$.



          The next term is fairly direct,
          $$biggllvert fraclbrace lvert trvertrbracelvert trvert^sbiggrrvert = fraclbracelvert trvertrbracelvert trvert^sigma < frac1lvert trvert^sigma < frace^100lvert trvert$$
          for $lvert trvert geqslant 1$ since $0 leqslant lbrace lvert trvertrbrace < 1$.



          For the integral term, we cannot go quite so far. The integral only makes sense for $operatornameRe s > 0$ (perhaps also for $operatornameRe s = 0$ and $sneq 0$, but not beyond that), so we must either restrict to $lvert trvert geqslant e^100$ or modify the lower bound for $sigma$ to
          $$sigma > max ; Biggl 0, 1 - frac100log lvert trvertBiggr,.$$
          Well, since the straightforward estimate of the integral,
          $$Biggllvert int_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant int_lvert trvert^infty fraclbrace wrbracew^1+sigma,dw leqslant int_lvert trvert^infty fracdww^1+sigma = frac1sigmacdot lvert trvert^sigma,,$$
          produces a $sigma$ in the denominator, we must keep $sigma$ away from zero, so we actually need a stronger restriction,
          $$sigma geqslant max; Biggl c, 1 - frac100log lvert trvertBiggr$$
          for a fixed $c in (0,1)$, or $lvert trvert geqslant t_0$ for a fixed $t_0 > e^100$.



          With that, we can estimate the last term by
          $$Biggllvert sint_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant fraclvert sigma + itrvertsigmalvert trvert^sigma leqslant fracsigma + lvert trvertsigmalvert trvert^sigma = lvert trvert^-sigma + fraclvert trvert^1-sigmasigma,.$$
          We had estimated $lvert trvert^-sigma$ above, and with that we obtain
          $$Biggllvert sint_lvert trvert^infty fraclbrace wrbracew^1+s,dwBiggrrvert leqslant frace^100lvert trvert + frace^100sigma leqslant e^100biggl(frac1lvert trvert + frac1cbiggr),.$$
          Thus each of the terms except the partial sum is bounded and we have
          $$zeta(sigma + it) = sum_n leqslant lvert trvert frac1n^sigma +it + O(1)$$
          in the indicated region. The remaining sum is estimated using
          $$lvert n^-sigma - itrvert = n^-sigma = fracn^1-sigman leqslant fraclvert trvert^1-sigman leqslant frace^100n$$
          for $sigma < 1$, and
          $$lvert n^-sigma-itrvert = n^-sigma leqslant frac1n$$
          for $sigma geqslant 1$.







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          answered Jul 26 at 14:35









          Daniel Fischer♦

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          • Thank you for such a detailed and clear response!
            – VBACODER
            Jul 26 at 19:32
















          • Thank you for such a detailed and clear response!
            – VBACODER
            Jul 26 at 19:32















          Thank you for such a detailed and clear response!
          – VBACODER
          Jul 26 at 19:32




          Thank you for such a detailed and clear response!
          – VBACODER
          Jul 26 at 19:32












           

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